I'm following https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowmore.pdf
and in particular the following paragraph:
Let $N = N_G(P)$ be the normalizer of $P$ in $G$. Then all the elements of $p$-power order in $N$ lie in $P$. Indeed, any element of $N$ with $p$-power order which is not in $P$ would give a non-identity element of $p$-power order in $N/P$. Then we could take inverse images through the projection $N\to N/P$ to find a $p$-subgroup inside $N$ properly containing $P$, but this contradicts the maximality of $P$ as a $p$-subgroup of $G$.
My question in the proof by contradiction.
$|z|=p^k$, $z\in N$ but $z\notin P$, then $nP\ne P$. Why is $nP$ an element of $p$-power? $(nP)^j=n^jP$, and thus $n^j=e$ only if $j$ is a p-power, but what about $n^j\in P$ with $j$ not p-power? Then we would have $|nP|=j$.
$\pi:N\to N/P$, since $nP\ne P$, $\pi^{-1}(nP)$ cannot be a group since $e$ does not belong to it. What is meant in the text? Perhaps $\langle \pi^{-1}(nP)\rangle$?
The answer to 2. is "Yes", what is meant is $\langle \pi^{-1}(nP)\rangle$.
For 1, I find it easiest to think about $\pi$ as just 'any' homomorphism from $N$ to 'some group' $Q$ (so we forget everything we know about $Q$ and $\pi$ except that $\pi$ is a homomorphism.) The claim now is that if $n \in N$ has $p$-power order in $N$, then so does $\pi(n)$ in $Q$.
In fact it is even worse: the order of $\pi(n)$ divides the order of $n$.
It all comes down to Euclid's algorithm from elementary number theory. Let $a$ be the order of $n$ in $N$ and $b$ be the order of $\pi(n)$ in $Q$. Using 'division with remainder' we have that $a = qb + r$ for some natural numbers $q$ and $r$ with $r < b$ (possibly $r = 0$).
Let $e$ be the unit of $Q$. Then $e = \pi(n^a) =\pi((n^b)^q n^r) = (\pi(n)^b)^q \pi(n)^r = e^q \pi(n)^r = e \pi(n)^r = \pi(n)^r$, or shorter: $\pi(n)^r = e$.
But since $b$ was the smallest positive number such that $\pi(n)^b = e$, and $r < b$ we have that $r = 0$ and hence $a = qb$ so that indeed $b|a$.